Question

The Sr-71 Blackbird whch is 107 ft 5 in long is the world's fastest airplane. it can fly three times the speed of sound (mach3) at altitudes of 80000 ft. When it lands after a long flight it is too hot to be touched for 30 minutes and is 6 inches longer than at takeoff. how hot is the blackbird when it lands assuming the coefficient of linear expansion is 2.4×10^-5 K^-1 and it's temperature at takeoff is 23°C?

Answer 1

When the temperature of an object increases or decreases, it experiences a change in length which is directly proportional to the initial length and temperature change. In other words,

dL ? Li * (t2 - t1),

Or, dL = k * Li * (t2 - t1).......................(1) , where

dL = change in length of the object , Li = initial length of the object , t1 = initial temperature of the object , t2 = final temperature of the object , k = co-efficient of linear expansion

So, as per question, after landing the airplane is 6 inches longer than at takeoff, hence dL = 6 in.

Initial length = Li = 107 ft 5 in = (1284 + 5) in = 1289 in (as 1 ft = 12 in)

Also, given that coefficient of linear expansion = k = 2.4×10^-5 K^-1

Temperature at takeoff (initial temperature) = t1 = 23°C = (273 + 23)K = 296K (as Kelvin temperature = 273 + Celsius temperature)

So, substituting these values in (1) above, we get:

6 = 1289 * 2.4×10^-5 * (t2 - 296) , solving which we get,

t2 ? 489.95 K (rounded off to 2 decimal places)

Or, t2 = (489.95 - 273)°C = 216.95°C

Thus, the blackbird is 216.95°C hot when it lands.